Question

The sides of triangle are in A.P. and the greatest angle exceeds the least by 90. The sides are in the ratio _____________.

• A. $1:2:\sqrt{2}$
• B. $1:\sqrt{3}:2$
• C. $\sqrt{7}+1:\sqrt{7}:\sqrt{7}-1$
• D. $\sqrt{3}+1:1:\sqrt{3}-1$

Answer 1

The correct option is C $\sqrt{7}+1:\sqrt{7}:\sqrt{7}-1$

If $a,b,ca,b,c$ where $a>b>c⟹a+c=2ba>b>c⟹a+c=2b$

$asin(90\circ +x)=bsin(90\circ -2x)=csinx=2Rasin(90\circ +x)=bsin(90\circ -2x)=csinx=2R$

$⟹a=2Rcosx,b=2Rcos2x,c=2Rsinx⟹a=2Rcosx,b=2Rcos2x,c=2Rsinx$

Using $a+c=2b,cosx+sinx=2cos2x=2(cosx-sinx)(cosx+sinx)a+c=2b,cosx+sinx=2cos2x=2(cosx-sinx)(cosx+sinx)$

As $90\circ -2x>0,cos2x>090\circ -2x>0,cos2x>0$

and also $cosx,sinx>0⟹cosx+sinx>0cosx,sinx>0⟹cosx+sinx>0$

cancelling $cosx+sinx,cosx+sinx$, we get $cosx-sinx=12cosx-sinx=12$

Squaring we get, $1-sin2x=14⟺sin2x=?,1-sin2x=14⟺sin2x=?$

$⟹cos2x=\sqrt{1-si{n}^{2}2x}⟹cos2x=+1-si{n}^{2}2x=\frac{7}{4}$

$cos2x=1+co{s}^{2}x2=4+7–\sqrt{8}=\sqrt{(7+14)}2cos2x=1+co{s}^{}2x2=4+78=(\sqrt{7}+14{)}^2$

Hence the ratio of side sre $\sqrt{7}+1:\sqrt{7}:\sqrt{7}-1$